\(\Rightarrow x^2-x+1=5x\Leftrightarrow x^2+1=6x\)
\(\dfrac{x^2}{x^4+x^2+1}=\dfrac{x^2}{\left(x^2+1\right)^2-x^2}=\dfrac{x^2}{\left(6x\right)^2-x^2}=\dfrac{1}{35}\)
\(\dfrac{x}{x^2-x+1}=\dfrac{1}{5}\left(1\right)\)
có x^2 -x+1 > 0 mọi x => x>0
\(\left(1\right)\Leftrightarrow\dfrac{x^2-x+1}{x}=5\)
\(x+\dfrac{1}{x}=6;x^2+\dfrac{1}{x^2}=34\)
\(\left(1\right)\Leftrightarrow\dfrac{1}{E}=\left(\dfrac{x^4+x^2+1}{x^2}\right)=x^2+1+\dfrac{1}{x^2}=34+1=35\)
\(E=\dfrac{1}{35}\)
Từ \(\dfrac{x}{x^2-x+1}=\dfrac{1}{5}\Rightarrow x^2-x+1=5x\Rightarrow x^2+1=5x+x=6x\)
Ta có: \(E=\dfrac{x^2}{x^4+x^2+1}\)
= \(\dfrac{x^2}{x^4-x+x^2+x+1}\)
= \(\dfrac{x^2}{x\left(x^3-1\right)+\left(x^2+x+1\right)}\)
= \(\dfrac{x^2}{x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)}\)
= \(\dfrac{x^2}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
= \(\dfrac{x}{x^2+x+1}.\dfrac{x}{x^2-x+1}\)
= \(\dfrac{x}{6x+x}.\dfrac{1}{5}\) = \(\dfrac{x}{5.7x}\) = \(\dfrac{1}{35}\)
